Question: $\text D = \left[\begin{array}{r}2 \\ 3 \\ 4\end{array}\right]$ and $\text A = \left[\begin{array}{rr}-2 & -2\end{array}\right]$ Let $\text {H = DA}$. Find $\text H$. $ {H = }$
The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{D}$ and the first column of $\text{A}$. $ \text {H}=\left[\begin{array}{rr}{2} \\ 3 \\ 4\end{array}\right]\left[\begin{array}{rr} {-2} & -2\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(2)\cdot(-2)\\\\ &=-4 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $3 \cdot -2 = -6$ (Choice B) B $2 \cdot -2 = -4$ (Choice C) C ${H}_{2,1}$ does not exist. Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}-4 & -4 \\ -6 & -6 \\ -8 & -8\end{array}\right] $